We can see from this picture, the y axis is a line of reflection for opposite values of b.

Also, clearly the y-intercept is -1 for all equations.

The chart below summarizes the nature of the roots for varying values of b.

b

Number of Roots

Nature of Roots

Vertex

-3

two

real, irrational

(1.5, -1.25)

-2

one

real, rational, x = 1

(1,0)

-1

two

complex, imaginary

(0.5, 0.75)

0

one

complex, imaginary, x = i

(0,1)

1

two

complex, imgainary

(-0.5 , 0.75)

2

one

real, rational, x = -1

(-1,0)

3

two

real, irrational

(-1.5, -1.25)

To determine the locus of vertices, one should notice the curve appears to be a parabola.

Step 1: To determine the equation of the quadratic curve, I chose three vertices (1,0), (0,1) and (-1,0) along the desired path.

Step 2: Using these three verices and the equation y = ax²+ bx + c, one obtain the resulting equations:

0 = a + b + c

1 = c

0 = a - b + c

Step 3: Thus letting c = 1, the two resulting equations are:

0 = a + b + 1

0 = a - b + 1

Step 4: Thus adding these two equations, you result with a value of a = -1. If a = -1, thus b = 0.

Step 5: One obtains the equation y = -x²+1 for the locus of vertices.

The following graphic shows the path of the parabolas from b = -3 to 3 and it's locus of vertices: y = -x²+1.

Now consider the xb plane. In this case we let y = b.

The following is the graph of x² + yx + 1 = 0 and the intersection of it with graphs of y =b, from b = -3 to 3

If we look at the table above for different values of b, we can see a pattern in the number of intersections each line makes with the equation 0 = x² + yx + 1

b

Number of Roots

Nature of Roots

Number of Intersection y = b with equation 0 = x² + yx + 1

-3

two

real, irrational

two

-2

one

real, rational, x = 1

one (1, -2)

-1

two

complex, imaginary

none

0

one

complex, imaginary, x = i

none

1

two

complex, imgainary

none

2

one

real, rational, x = -1

one (-1, 2)

3

two

real, irrational

two

We notice when b = -2, the root is 1, which is also the x value for the intersection of y = -2 with the equation 0 = x² + yx + 1.

We notice when b = 2, the root is -1, which is also the x value for the intersection of y = 2 with the equation 0 = x² + yx + 1.

We notice when b = -1, 0, or 1, the roots are imaginary and y = b does not intersect with the equation 0 = x² + yx + 1.

As we look to b = -3 and 3 one notices there are two roots for the original y = x² + bx + 1. I conjecture that the value of the roots for each value of b will be the x value of the intersection of the graph y = b and 0 = x² + yx + 1. When solving for the roots of the equation y = x² -3x + 1, using the quadratic formula, one obtains the values x = (3 ±√5)/2, and for y = x² +3x + 1, one obtains the values

x = (-3 ±√5)/2, which are the values of the intersection with the graph y = b.